# Tag Info

## Hot answers tagged thermal-field-theory

9

Here is a proof following Ojima, "Lorentz Invariance vs. Temperature in QFT", Letters in Mathematical Physics (1986) Vol. 11, Issue 1 (1986) 73-80. The first two pages of the paper are available for free here, but the website wants money for more of the paper. (Click the orange "Look Inside" button if the paper doesn't open automatically.) Fortunately, the ...

6

We calculate the free energy (density) for the Higgs field $\phi$ at finite temperature. In the Standard Model, this looks like $\mathcal{F}_{SM}(\phi,T) = -\frac{\pi^2}{90}g_* T^4+V_{SM}(\phi, T) \ ,$ where $g_*$ is the number of degrees of freedom in the SM ($g_*=106.75$). The potential has the form $V_{SM}(\phi,T) = D(T^2-T_0^2)\phi^2 - ET\phi^3+\frac{... 5 An enclosed cavity near thermal equilibrium at some temperature will be filled with blackbody radiation, which has a well-defined spectrum that depends only (!) on the temperature of the cavity. If you poke a small hole in the cavity, some of those photons will leak out of the hole as thermal radiation. For instance, the pupils of your eyes are dark ... 4 There are two particle candidates for dark matter, and both are still in the realm of hypothesis. These particles are candidates exactly because they cannot decay to something lighter, they are stable. Neutrinos, which also cannot decay to something lighter, has such a small mass that they cannot model the way dark matter is attracted to gravitational ... 3 You're using different boundary conditions around the thermal circle, so there's no reason to expect the results to be related. When you use periodic boundary conditions around the thermal circle, it's equivalent to inserting a factor$(-1)^F$in the trace $$Z = {\rm Tr\ } (-1)^F e^{-\beta H}$$ and because of supersymmetry, only the groundstates can ... 3 That trace is nothing but the sum of the non-vanishing$\rho$eigenvalues to the 1/r power and taking their multiplicities into account. Since those eigenvalues$\lambda$belong to$(0,1]$, then$\lambda^{1/r}\to 1$as$r\to +\infty$. In summary$tr(\rho^{1/r})$tends to the number of non-vanishing eigenvalues of$\rho$taking multiplicities into account. In ... 3 That is the same situation as in QFT, as soon as you drop Poincaré invariance. I mean in curved spacetime. There you may have uncountably many inequivalent representations of the same algebra of observables just by varying some continuous parameter (curvatures). If you put all these reps in orthogonal sectors in a common Hilbert space, it must be non.... 3 Roughly speaking solid matter is on a lattice form, A three-dimensional lattice filled with two molecules A and B, here shown as black and white spheres. The molecules fit like LEGO , the forces tying them together are mainly the spill over electric field forces , attractive and repulsive forming the patterns of the lattice. In a single crystal one ... 2 1) What you call$\rho$should really be called$\epsilon$(this is the energy density, not the particle density). 2) The thermodynamic variables$\epsilon$and$P$are expectation values of certain operators in a thermal ensemble. You should not confuse equations for the operators with equations for thermodynamic quantities. 3) The operator that ... 2 There seems to be some confusion in your question between thermal fluctuations and quantum fluctuations, so I will try to address both of them in my answer. Spontaneous symmetry breaking occurs when the ground state of the Hamiltonian does not exhibit the full symmetries of the Hamiltonian. In other words, there are multiple degenerate configurations with ... 2 The point is that to achieve the sum over Matsubara frequencies $$\sum_{n} g(i\omega_n)$$ we can use a contour integral $$\oint_C g(z) f(z)$$ with the contour described in fig 1 here, so long as we choose an$f(z)$with simple poles exactly at the Matsubara frequencies$\omega_n$. This determines$f(z)$to be proportional to the Bose-Einstein distribution. ... 2 A Euclidean correlation function may be interpreted as a Lorentzian expectation value by "cutting" the path integral and continuing the time coordinate. Let me review how this procedure relates Euclidean correlation functions on a closed manifold$M\times S^1_\beta$to thermal expectation values in a Lorentzian quantum field theory on$M \times \mathbb{R}...

2

Two remarks: Essentially every function can be regarded as a distribution, although the converse is not true. In this sense, you can indeed regard $(e^{\beta |p_0|} - 1)^{-1}$ as a distributions. Distributions of this kind are known as regular distributions. Non-regular distributions are known as singular distributions. It is not true that you cannot ...

2

First write $\rho, y$ in the eigenbasis of the Hamiltonian, $$\rho = \sum_n p_n |n\rangle \langle n| \\ y = \sum_n q_n |n\rangle \langle n|$$ with $q_n^4 = p_n$, in particular $0\le q_n \le 1$. In this basis the inequality becomes $$\sum_{n,m} |V_{n,m}|^2 q_n^{1+\eta} q_m^{3-\eta} \le \sum_{n,m} |V_{n,m}|^2 q_n q_m^2$$ At this point the ...

1

I am not going to go into the black hole stuff, but thermalization is never loss of matter and/or probability, but - so to say - only a "rearrangement" of density matrix eigenstates and corresponding probabilities. In other words, it is an evolution from an arbitrary initial state $\rho$, which may be pure ($\;\rho = \rho^2 = |\Psi\rangle\langle \Psi |\;$) ...

1

The effective action is defined so that $\frac{\delta \Gamma \left[ \phi \right]}{\delta \phi^i \left( x \right)}=J_{\phi}^i \left( x \right)$, where $J$ is a classical current, so we see that this definition coincides with the usual definition of the action in classical field theory. Moreover, the procedure of evaluation of this action involves evaluation ...

1

Concerning OP's explicit example $$\frac{\delta(p^2 + m^2 )}{e^{\beta |p_0|} - 1} , \qquad\beta\neq 0. \tag{D}$$ The massive case $m\neq 0$. Then the singularity of $\frac{1}{e^{\beta |p_0|} - 1}$ does not overlap with the support of $\delta(p^2 + m^2 )$, so the product distribution is mathematically well-defined. The massless case $m=0$. Then the ...

1

No. Consider for example observables of the form $\mathcal O = e^{\beta H/2}$; the value exponentially increases as the probability exponentially decreases and you can get what's in principle an infinite sum of constant terms.

1

In general, the expectation value $tr(A e^{-\beta H})$ is not defined for a generic selfadjoint operator $A$ (of the form ${\cal O}^2$ or not) if it is unbounded as it is the standard situation in QFT. $tr(A e^{-\beta H})$ however converges if (with the written order!) the range of $e^{-\beta H}$ belongs to the domain of $A$ and the composition is trace ...

1

I think the second formula is also not very profound. This is just based on the fact that the chemical potential $$\Delta S = \int dt \, \mu Q$$ enters the action like the zeroth component of an (imaginary) $U(1)$ gauge field $A_\beta=(i\mu,\vec{0})$. Note that this is the Polyakov line for a $U(1)$ background field, not the Polyakov line of (for example)...

1

Temperature is a macroscopic measure, it is statistical , a thermodynamic variable. At best, an ensemble of particles with a statistical distribution of energy in a thermodynamic equilibrium will have a kinetic temperature defined as connected with the average kinetic energy of the ensemble of particles. On the other hand, the scattering cross section ...

1

You can always go straightforward and use that you know how $\hat{a}$ and $\hat{a}^\dagger$ act on the basis $|n\rangle$. Let's write down straightforwadly what $\mathrm{Tr}$ is $$\mathrm{Tr}\Big(\hat{O}e^{-\beta\hat{H}}\Big)=\sum_n\langle n|\hat{O}e^{-\beta\hat{H}}|n\rangle=\sum_n e^{-\beta(n+1/2)}\langle n|\hat{O}|n\rangle$$ Now,...

1

Yes emissivity depends on temperature: $$\epsilon(T)= \frac{E(T)}{E_b (T)}$$ $\epsilon$ is total hemespherical emissivity. $E$ is the emissive power of the actual body which depend on temperature and $E_b$ is the emissive power of a blackbody: $E_b(T)=\sigma T^4$

1

Collisions of a state with other particles, present at finite density, influence the life-time of the state as energy can be transferred to those other particles during inelastic collisions (i.e. they can change state). This change in life-time is related to a change in the width via the uncertainty relation and gives the thermal width. There is a physical ...

1

Finite temperature Feynman rules are simply zero temperature Feynman rules for Euclidean ($t\to i\tau$) QFT in periodic imaginary time. So instead of continuous values for the momenta, you will have a discrete spectrum for the timelike moments (such as in the infinite potential well in basic quantum mechanics). It's called the Matsubara formalism, if you ...

1

Thermal expansion from an atomistic perspective: The energetic potential between two atoms can be approximated by two exponential functions, one for the attractive force between the atoms, one for the repulsive force. The superposition of these two force fields has a minimum at a certain distance. Examples for such empirical potentials are Stillinger-Weber, ...

1

Let's define $T_{EW}$ the temperature where the coefficient $m^2_H(T)$ of the operator $H^2$ in the SM lagrangian vanishes: $$m_H^2(T=T_{EW})=0\,.$$ For $T>T_{EW}$ the Higgs vev is vanishing, the EW symmetry in unbroken, and the elementary particles are massless. For $T<T_{EW}$ the the vev is non-vanishing, $v_T\propto -m_H^2/\lambda\neq 0$, the EW ...

1

A simple (and quite accurate) answer is that quantum fluctuations are the fluctuations that exist at zero temperature. What it means is that even at zero temperature, there might be fluctuations in the measurements of observables, which does not happen for classical systems at zero temperature, due to the non-commutativity of the dynamical and potential ...

1

Found a sketch of a proof on a referee's report on a paper RELATIVISTIC INVARIANCE OF THE VACUUM by Adam Bednorz. The referee's sketch is: Comment Hundreds of calculations in Fnite temperature Feld theory have been published. To my knowledge, none of these calculations have ever conflicted with Lorentz invariance in the limit $\beta \to \infty$ ...

Only top voted, non community-wiki answers of a minimum length are eligible